Name: tangent plane estimates for two variables
Uploaded: 2026-06-08T09:38:27.256Z
Duration: 1 min 23 s
Channel: JoVE

假设某城市中的污染物浓度被建模为关于 x 和 y 的可微函数，并由 xy 平面上方的一个三维曲面表示。由于浓度变化是连续的，因此可以根据已知的测量值来估算邻近位置的浓度值。此过程称为线性近似。它利用局部平面模型来估算特定点（例如污染物浓度已知的监测站）附近的污染水平。尽管污染物表面可能在城市范围内呈曲面分布，但在靠近该监测站的局部区域，其形态可近似为一个平面。这一平面称为切平面，由已知的污染物浓度以及在 x 和 y 方向上的偏导数共同定义。因此，使用切平面而非曲面来估算邻近位置的浓度。当位置非常接近监测站时，该估算通常较为准确。但随着位置逐渐远离，该近似结果的准确性会降低，因为污染曲面不再接近于该平面模型。通过这种方式，线性近似可对复杂曲面提供快速的局部估计。

线性近似

对于二元可微函数，线性近似通过用其切平面替代曲面，来估计已知点附近的函数值。考虑以下函数\begin{equation*}f(x,y)=x^2+3y^2\end{equation*}在点 (2, 1) 附近。该点的精确值为 f(2, 1) = 22 + 3(1)2 = 4 + 3 = 7。函数 f(x, y) 在点 (a, b) 附近的线性近似为\begin{equation*}L(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)\end{equation*}首先，计算偏导数：fx(x, y) = 2x 和 fy(x, y) = 6y。在点 (2, 1) 处，fx(2, 1) = 4 且 fy(2, 1) = 6。因此，线性近似为\begin{equation*}L(x,y)=7+4(x-2)+6(y-1)\end{equation*}为了估算 f(2.1, 0.9)，代入 x = 2.1 和 y = 0.9：\begin{equation*}L(2.1,0.9)=7+4(2.1-2)+6(0.9-1) = 6.8\end{equation*}线性近似估计 f(2.1, 0.9) 约为 6.8。其精确值为\begin{equation*}f(2.1,0.9)=(2.1)^2+3(0.9)^2=4.41+2.43=6.84\end{equation*}因此，由于 (2.1, 0.9) 接近 (2, 1)，切平面估计值接近实际值。

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Education

教育

JoVE Core

Calculus

微积分

Unit

单元

Partial Derivatives and Gradients

偏导数与梯度

KEY TERMS AND CONCEPTS

关键术语与概念

Linear Approximations

tangent plane estimates for two variables

Tangent plane estimates help approximate a differentiable function of two variables near a known point. Instead of using the curved surface directly, the function is replaced by its tangent plane. This gives a fast estimate for values close to the point being studied.

For the function f(x,y)=x^2+3y^2, the point (2, 1) is used as the reference. The exact value there is f(2,1)=2^2+3(1)^2=4+3=7. The linear approximation near (a,b) is L(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b).

To build the estimate, the partial derivatives are found first. For this function, f_x(x,y)=2x and f_y(x,y)=6y. At (2,1), the partial derivatives are f_x(2,1)=4 and f_y(2,1)=6, so the tangent plane formula becomes L(x,y)=7+4(x-2)+6(y-1).

This plane can be used to estimate nearby values. For example, substituting x=2.1 and y=0.9 gives L(2.1,0.9)=7+4(2.1-2)+6(0.9-1)=6.8. The exact value is f(2.1,0.9)=(2.1)^2+3(0.9)^2=4.41+2.43=6.84, so the estimate is very close because the point is near (2,1).

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Use tangent plane estimates to approximate a two-variable function near a known point, find partial derivatives, and compare the estimate with the exact value.

Tangent Plane Estimates for Two Variables | JoVE K12

The JoVE Core Calculus collection provides a comprehensive visual guide to the entire calculus sequence, seamlessly connecting mathematical theory to real-world applications in physics and economics. Through dynamic animated lessons, the course transitions smoothly from foundational single-variable concepts to advanced multivariable topics. Students will explore everything from standard differentiation and integration to spatial vectors, multi-dimensional derivatives, and complex vector fields. By breaking down abstract coordinate systems, infinite series, and spatial motion, this structured approach builds the robust analytical skills and advanced mastery essential for STEM success.

calculus

Tangent Plane Estimates for Two Variables

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